Easiest way in general to find the sin, cos, arcsin, arccos, of "not so easy" angles/values without using a calculator? I was wondering if there are any easy ways in general to find the sin, cos, arcsin, arccos, of "not so easy" angles/values without using a calculator. By "not so easy" I mean just not things like 0, $\pi/6$, $\pi/4$, $\pi/3$, or $\pi/2$, which one encounters routinely. And also not $\sin(\theta+2n\pi)$, $\sin(n\pi)$, $\sin(-\theta)$, and the like. The reason I'm asking this terribly general question (please forgive me for that) is because in Calculus class I always make some sort of error when these sorts of calculations need to be done to solve a problem.
For example, on a recent question I had to find $0<\theta<2\pi$ satisfying $\sin\theta=-1/2$. 
We know $\sin(\pi/6)=1/2$ and $\sin(-\pi/6)=-1/2$ since $-\sin(x)=\sin(-x)$. Also we know sin has period $2\pi$ so $\sin(-\pi/6+12\pi/6)=\sin(11\pi/6)$=$-1/2$. 
But how would one get $\sin(7\pi/6)=-1/2$ for example?
Or similarly how would one know the value of sin $4\pi/3$, $5\pi/3$, $5\pi/4$, $6\pi/5$, $5\pi/6$, etc. Or cos, arcsin, arccos, of such "nasty" values?
Again, I apologize for the ridiculously general question, it's just that I know no other way of asking it.
 A: For the simpler question (the only ones you'll have to deal with in introductory calculus classes) of how to figure out multiples of $\pi/6$ and $\pi/4$, imagine the angles as being created by points on a unit circle.
Now, find the coordinates of each point: if it's a multiple of $\pi/4$, you know that the side lengths of a right isosceles triangle are $1:1:\sqrt{2}$, and if it's a multiple of $\pi/6$, you know that the side lengths of half an equilateral triangle are $1:2:\sqrt{3}$.
Divide the y-coordinate by the x-coordinate and you have the tangent; divide the y-coordinate by the hypotenuse and you have sin; divide the x-coordinate by the hypotenuse and you have cos.

A: Since $\pi/4$ is an angle of a $1:1:\sqrt{2}$ right-angled triangle, and $\pi/6$ and $\pi/3$ are angles of a $1:\sqrt{3}:2$ right-angled triangle, the $\sin / \cos / \tan$ of these and their multiples are easy to calculate on the fly.
$\sin 6\pi/5$  or something similar would be harder.
A: Lots of usage of various formula related to $\sin nx$ and $\cos nx$ and half-angle formulae. 
For example, if $x=\pi/5$ then $0=\sin \pi = \sin 5x$. You can expand: $$\sin 5x= \sin x\left(16\cos^4 x - 12\cos^2 x+1\right)$$
So $y=\cos x$ is a root of $16y^4-12y^2+1=0$. We can solve for $y^2$ using the quadratic formula, then take the square root to get $y$. (You actually end up with four potential solutions, but you can see what they represent.)
In general, we know that $\sin(\pi +x)=-\sin(x)$ and $\cos(\pi +x)=-\cos(x)$. So you can always reduce the problem to $0\leq x<\pi$.
Use $\cos (A+B)=\cos A \cos B-\sin A\sin B$ to come up with a formula for $\cos nx$ and $\sin nx$. Look up Chebyshev polynomials if you want the advanced treatment.
